Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is spin? (advanced question)

  1. Jun 26, 2006 #1
    I had thought that if under an infinitesimal rotation by [tex]\theta[/tex] around the z-axis a quantity transforms according to

    [tex]\psi\rightarrow\left(1+i\left[S+\left(\frac{1}{i}x\frac{\partial}{\partial y}-\frac{1}{i}y\frac{\partial}{\partial x}\right)\right]\theta\right)\psi[/tex]

    then the operator S is by definition the z-component of the spin operator for that quantity.

    However, the Dirac field operator transforms (in the appropriate representation) according to

    [tex]\psi\rightarrow\left(1+i\left[\frac{1}{2}\left(\begin{array}{cc}\sigma_{z}&0\\ 0&\sigma_{z}\end{array}\right)+\left(\frac{1}{i}x\frac{\partial}{\partial y}-\frac{1}{i}y\frac{\partial}{\partial x}\right)\right)\theta\right]\right)\psi[/tex]

    yet https://www.amazon.com/gp/product/0521478146/103-7918027-3072619?v=glance&n=283155 states that [tex]\tau_{z}=\frac{1}{2}\left(\begin{array}{cc}\sigma_{z}&0\\0&\sigma_{z}\end{array}\right)[/tex] cannot be the spin operator because it does not commute with [tex]\gamma^{\mu}\partial_{\mu}[/tex] The correct spin operator is--or has something to do with--the Pauli-Lubanski pseudovector.

    My question is, first, what's wrong with not commuting with [tex]\gamma^{\mu}\partial_{\mu}[/tex]? Why is that necessary for the a good spin operator?

    And secondly, if [tex]\tau_{z}[/tex] is not a good spin operator, what about the fact that it generates the non-scalar part of a rotation? Is that not the definition of spin?
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Jan 2, 2007 #2
    I don't know a definitive answer. However, isn't a part of [itex]\gamma^{\mu}\partial_{\mu}[/itex] the Hamiltonian? And (of course) if we want the eigenvalues of some operator to be constant in time, we require the operator to commute with the Hamiltonian.
  4. Jan 3, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    The spin needn't be conserved. Actually for the Dirac Hamiltonian it isn't. It's only the total angular momentum that is conserved for a Hamiltonian generated time-evolution.

    Actually, besides the total momentum and angular momentum, the Pauli-Lyubanski pseudovector also commutes with the Hamiltonian and therefore is conserved for all projective unitary irreds of the restricted Poincare' group.

  5. Jan 4, 2007 #4


    User Avatar
    Science Advisor

    I have also problems to understand that section of Ryder's book. In the part where Wigner's little group is discussed it is shown in detail that the non-relativistic spin operator is the correct spin operator in the rest frame of a particle. Moreover, it is shown how the Pauli-Lubanski pseudovector reduces basically to the non-relativistic spin operator in the rest frame. But it seams to me that he does not spend a word to explain why the Pauli-Lubanski pseudovector should actually provide a generalized relativistic spin operator.
    Last edited: Jan 5, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook